SOLUTION: if the sum of the digits of a 3 digit number is divisible by 9, prove that the 3 digit number is also divisible by 9

Question 995573: if the sum of the digits of a 3 digit number is divisible by 9, prove that the 3 digit number is also divisible by 9
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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Let us consider an arbitrary three-digit number "abc".

This three-digit number "abc" is   100*a + 10*b + c.   (It is how our decimal system works).

Now, regroup it:   100*a + 10*b + c = (99*a+a) + (9*b+b) + c = (99*a + 9*b) + (a + b + c).

The additive (99*a + 9*b) is divisible by 9. Therefore, the divisibility by 9 of our original number "abc" depends on and is determined solely by the last additive (a + b + c), which is the sum of the digits of the number "abc".

This "divisibility by 9" rule is true not only for 3-digit numbers: it is true for all integer numbers.

For the proof see the lesson Divisibility by 9 rule in this site.

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